Definitions of Differential Equations

Differential Equations: Equations that involve derivatives

Examples:dxdt+x=2cos(t)d2xdt2+dxdt−x=0\text{Examples:}\\ \frac{dx}{dt} + x = 2 cos(t) \\ \frac{d^2 x}{d t^2} + \frac{dx}{dt} - x = 0

In these examples, x is dependent on t; can be considered a function. On the other hand, t is an independent variable. Differential equations can also include parameters which are constants/coefficients.

Is x=cost+sint a solution to dxdt+x=2cost?dxdt+x=ddt(cost+sint)+cost+sint=(−sint+cost)+cost+sint=2costThus, x is a solution to the equation, and there are many other x that are solutions as well.\text{Is } x = cost + sint \text{ a solution to } \frac{dx}{dt} + x = 2 cost \text{?} \\ \frac{dx}{dt} + x = \frac{d}{dt} (cost + sint) + cost + sint = (-sint + cost) + cost + sint = 2cost \\ \text{Thus, x is a solution to the equation, and there are many other x that are solutions as well.}
  • Ordinary differential equation: A differential equation with only one independent varialbe
    • $\frac{dx}{dt} = sin x$
  • Partial differential equation: A differential equation with several independent variables, using partial derivatives
    • $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 1$
  • Linearity: An equation is linear if and only if the dependent variables and their derivatives appear linearly
    • Linear: $\frac{dx}{dt} + t = 0 \text{ and } \frac{dx}{dt} + sin t = 0$
    • Nonlinear: $e^{dx / dt} = 0 \text{ and } e^x \frac{dx}{dt} + t = 0$
  • Homogeneity: A linear equation is homogeneous if all terms depend on the dependent variable. Otherwise, it is non- or inhomogeneous
    • Homogeneous: $\frac{d^2x}{dt^2} + e^t \frac{dx}{dt} + x = 0$
    • Inhomogeneous: $\frac{d^2x}{dt^2} - 2 \frac{dx}{dt} + t = 0$
  • Constant Coefficients: A linear equation has constant coefficients if the coefficients are all constants (except for the inhomogeneity)
    • Example: $5\frac{d^3x}{dt^3} + 4\frac{dx}{dt} + sin(t)cos(t) = 0$
  • Autonomy: An equation is autonomous if the equation does not depend on the independent variable at all
    • Autonomous: $e^x\frac{dx}{dt} + x = 0$
    • Not autonomous: $\frac{d^2x}{dt^2} + sint = 0$

Solving First Order ODEs

Case 1: f only depends on x

Given an equation like y′=f(x), we can use the anti-derivative to solve.∫y′dx=∫f(x)dxy(x)=∫f(x)dx+CWe can get rid of the constant given an initial value for y(c) where c is some constant; known as an IVP.\text{Given an equation like } y' = f(x) \text{, we can use the anti-derivative to solve.} \\ \int y'dx = \int f(x)dx \\ y(x) = \int f(x)dx + C \\ \text{We can get rid of the constant given an initial value for y(c) where c is some constant; known as an IVP.}

Case 2: f only depends on y

Equation: y′=f(y)Note that y′=dydx=f(y). Therefore:dydx=f(y)→dxdy=1f(y)Then, we can use the same method as Case 1, but we set x as the dependent variable and then solve for y.\text{Equation: } y' = f(y) \\ \text{Note that } y' = \frac{dy}{dx} = f(y) \text{. Therefore:} \\ \frac{dy}{dx} = f(y) \rightarrow \frac{dx}{dy} = \frac{1}{f(y)} \\ \text{Then, we can use the same method as Case 1, but we set x as the dependent variable and then solve for y.}
  • Can also be morphed into case 3

Case 3: f is depends on x and y (Separable)

Equation: y′=h(x)g(y)dydx=h(x)g(y)dyg(y)=h(x)dx∫1g(y)dy=∫h(x)dx+CNote that we might not get a clean definition of y. We can leave y in a more complex form as an implicit solution.‾\text{Equation: } y' = h(x)g(y) \\ \frac{dy}{dx} = h(x)g(y) \\ \frac{dy}{g(y)} = h(x)dx \\ \int \frac{1}{g(y)} dy = \int h(x)dx + C \\ \text{Note that we might not get a clean definition of y. We can leave y in a more complex form as an \underline{implicit solution.}} Example:dydx=xydyy=xdx∫1ydy=∫xdxln∣y∣=12x2+C∣y∣=e12x2+Cy=±eCe12x2→y=Ce12x2\text{Example:} \\ \frac{dy}{dx} = xy \\ \frac{dy}{y} = xdx \\ \int \frac{1}{y} dy = \int xdx \\ \text{ln} \vert y \vert = \frac{1}{2}x^2 + C \\ \vert y \vert = e^{\frac{1}{2}x^2 + C} \\ y = \pm e^C e^{\frac{1}{2}x^2} \rightarrow y = Ce^{\frac{1}{2}x^2} Newton’s Cooling Law Example:T(0)=89,T(1)=85,A=22dTdt=−k(T−A)=−kT+kA=−kT+22k1−k(T−22)dT=dt−t+C=1kln∣(T−A)∣∣(T−A)∣=e−kt+C=Ce−ktk(T−A)=Ce−ktGeneral solution: T=Ce−kt+A89=C+22→C=6785=67e−k+22→k=−ln(63/67)T(t)=67eln(63/67)t+2260=67eln(63/67)t+22→t≈9.21 minutes\text{Newton's Cooling Law Example:} \\ T(0) = 89, T(1) = 85, A = 22 \\ \frac{dT}{dt} = -k(T - A) = -kT + kA = -kT + 22k \\ \frac{1}{-k(T - 22)}dT = dt \\ - t + C = \frac{1}{k} ln \vert (T - A) \vert \\ \vert (T - A) \vert = e^{-kt + C} = Ce^{-kt} \\ k(T - A) = Ce^{-kt} \\ \text{General solution: } T = Ce^{-kt} + A \\ 89 = C + 22 \rightarrow C = 67 \\ 85 = 67e^{-k} + 22 \rightarrow k = - ln(63/67) \\ T(t) = 67e^{ln(63/67)t} + 22 \\ 60 = 67e^{ln(63/67)t} + 22 \rightarrow t \approx 9.21 \text{ minutes}

First Order Linear ODEs

Given the following sample linear ODE: a1(x)y′+a0(x)y=b(x)\text{Given the following sample linear ODE: } \\ a_1(x)y' + a_0(x)y = b(x) \\
  • If b(x) = 0, then the equation is homogenous
  • If b(x) != 0, then the equation is inhomogeneous
  • When $a_1(x) \neq 0$, then we can simplify this as $y’ + \frac{a_0(x)}{a_1(x)}y = \frac{b_0(x)}{a_1(x)} \rightarrow y’ + p(x)y = f(x)$
    • If p(x) = 0, then $y’ = f(x)$ which is the same as Case 1
    • If f(x) = 0, then $y’ + p(x)y = 0$ which can be written as a separable equation; Case 3
  • If p and f are nonzero, then $y’ + p(x)y = f(x)$
y′+p(x)y=f(x)To solve, we want to find a factor r(x) such that r(x)(y′+p(x)y)=r(x)f(x)→ddx(r(x)y)r(x)y=∫r(x)f(x)dx+Cy=r(x)−1(∫r(x)f(x)dx+C)Finding r(x):r(x)(y′+p(x)y)=ddx(r(x)y)=drdxy+r(x)dydxr(x)p(x)y=drdxy→drdx=r(x)p(x) (Separable equation)1rdr=p(x)dxr=Ce∫p(x)dx, AKA the integrating factory' + p(x)y = f(x) \\ \text{To solve, we want to find a factor r(x) such that } r(x)(y' + p(x)y) = r(x)f(x) \rightarrow \frac{d}{dx} (r(x)y)\\ r(x)y = \int r(x)f(x)dx + C \\ y = r(x)^{-1} (\int r(x)f(x)dx + C) \\ \text{Finding r(x):} \\ r(x)(y' + p(x)y) = \frac{d}{dx}(r(x)y) = \frac{dr}{dx}y + r(x)\frac{dy}{dx} \\ r(x)p(x)y = \frac{dr}{dx}y \rightarrow \frac{dr}{dx} = r(x)p(x) \text{ (Separable equation)} \\ \frac{1}{r} dr = p(x)dx \\ r = C e^{\int p(x)dx} \text{, AKA the integrating factor}

Examples

y′+2xy=ex−x2r(x)=e∫2xdx=ex2ex2(y′+2xy)=ddx(ex2y)=ex2ex−x2=exex2y=ex+C→y=ex−x2+Cex2 (General Solution)−1=1+C1→C=−2y=ex−x2−2ex2 (Particular Solution)y' + 2xy = e^{x-x^2} \\ r(x) = e^{\int 2xdx} = e^{x^2}\\ e^{x^2} (y' + 2xy) = \frac{d}{dx} (e^{x^2}y) = e^{x^2}e^{x-x^2} = e^x \\ e^{x^2}y = e^x + C \rightarrow y = e^{x - x^2} + \frac{C}{e^{x^2}} \text{ (General Solution)} \\ -1 = 1 + \frac{C}{1} \rightarrow C = -2 \\ y = e^{x - x^2} - \frac{2}{e^{x^2}} \text{ (Particular Solution)} 1xdydx−2yx2=xcos(x)dydx−2xy=x2cos(x)r(x)=e∫p(x)dx=e−2ln∣x∣=eln(∣x∣)−2=1x2ddx[1x2y]=1x2x2cos(x)=cos(x)1x2y=sinx+C→y=x2(sin(x)+C)\frac{1}{x} \frac{dy}{dx} - \frac{2y}{x^2} = x cos(x) \\ \frac{dy}{dx} - \frac{2}{x}y = x^2 cos(x) \\ r(x) = e^{\int p(x)dx} = e^{-2 ln \vert x \vert} = e^{ln (\vert x \vert)^{-2}} = \frac{1}{x^2} \\ \frac{d}{dx} [\frac{1}{x^2} y] = \frac{1}{x^2} x^2 cos(x) = cos(x) \\ \frac{1}{x^2} y = sin x + C \rightarrow y = x^2(sin(x) + C) \\

Substitution

Example: y′=(x+y+1)2Let u=x+y+1→u(x)=x+y(x)+1dudx=u′=(x+y+1)′=1+y′y′=u′−1New equation: u′−1=u2→dudx=u2+1Degenerates into case 2: dxdu=1u2+1x=∫1u2+1du+Cx=arctan(u)+C→u=tan(x−C)=tan(x+C)y=tan(x+C)−x−1\text{Example: } y' = (x + y + 1)^2 \\ \text{Let } u = x + y + 1 \rightarrow u(x) = x + y(x) + 1 \\ \frac{du}{dx} = u' = (x + y + 1)' = 1 + y' \\ y' = u' - 1 \\ \text{New equation: } u' - 1 = u^2 \rightarrow \frac{du}{dx} = u^2 + 1 \\ \text{Degenerates into case 2: } \frac{dx}{du} = \frac{1}{u^2 + 1} \\ x = \int \frac{1}{u^2 + 1} du + C \\ x = arctan(u) + C \rightarrow u = tan(x - C) = tan(x + C) \\ y = tan(x + C) - x - 1
  • Possible substitutions
Form Method Replacement
$yy’$ $u = y^2, u’ = 2yy’$ $yy’ = \frac{1}{2}u’$
$y^2y’$ $u = y^3, u’ = 3y^2y’$ $y^2y’ = \frac{1}{3}u’$
$cos(y)y’$ $u = sin(y), u’ = cos(y)y’$ $cos(y)y’ = u’$
$sin(y)y’$ $u = cos(y), u’ = -sin(y)y’$ $sin(y)y’ = -u’$
$e^yy’$ $u = e^y, u’ = e^yy’$ $e^yy’ = u’ $
2yy′+1=y2+xLetu=y2u′=2yy′u′+1=u+x→u′−u=x−1ddx[e−xu]=e−x(x−1)e−xu=∫e−x(x−1)dx+C=−e−x(x−1)−∫−e−xdx+C=−e−x(x−1)−e−x+Cu=−(x−1)−1+Cex=−x+Cexy=±(−x+Cex)0.52yy' + 1 = y^2 + x \\ \text{Let} u = y^2 \\ u' = 2yy' \\ u' + 1 = u + x \rightarrow u' - u = x - 1 \\ \frac{d}{dx} [e^{-x} u] = e^{-x} (x-1) \\ e^{-x} u = \int e^{-x}(x-1)dx + C \\ = -e^{-x}(x-1) - \int -e^{-x} dx + C \\ = -e^{-x}(x-1) - e^{-x} + C \\ u = -(x-1) - 1 + Ce^x = -x + Ce^x \\ y = \pm(-x + Ce^x)^{0.5}
  • Bernoulli equation is the first-order ODE of the form $y’ + p(x)y = q(x)y^n$
    • For n > 2, this equation would not be linear
    • Strategy: multiply the entire equation by $y^{-n}$ to turn it into $y^{-n}y’ + p(x)y^{1-n} = q(x)$
      • Substitute using $u = y^{1-n}, u’ = (1-n)y^{-n}y’$

Homogenous Equations

  • Homogenous equations can be given by the form $\frac{dy}{dx} = F(\frac{y}{x})$
  • Can be solved using a substitution $u = \frac{y}{x}$
    • $\frac{dy}{dx} = F(\frac{y}{x})$ becomes $u + xu’ + F(u)$
  • This new equation is separable; $\frac{1}{F(u) - u}du = \frac{1}{x}dx$

Example

x2y′=y2+xy, y(1)=1y′=y2x2+yxu=yx,y′=u+xu′u+xu′=u+u2xu′=u2u−2du=x−1dx−1u=ln∣x∣+Cu=−1ln∣x∣+Cyx=−1ln∣x∣+Cy=−xln∣x∣+C1=−1C→C=−1y=−xln∣x∣−1x^2y' = y^2 + xy \text{, } y(1) = 1 \\ y' = \frac{y^2}{x^2} + \frac{y}{x} \\ u = \frac{y}{x}, y' = u + xu' \\ u + xu' = u + u^2 \\ xu' = u^2 \\ u^{-2}du = x^{-1}dx \\ -\frac{1}{u} = ln \vert x \vert + C \\ u = - \frac{1}{ln \vert x \vert + C} \\ \frac{y}{x} = - \frac{1}{ln \vert x \vert + C} \\ y = - \frac{x}{ln \vert x \vert + C} \\ 1 = - \frac{1}{C} \rightarrow C = -1 \\ y = - \frac{x}{ln \vert x \vert -1}

Autonomous Equations

  • Autonomous equations are given by $y’ = f(y)$
  • A critical point of $x’ = f(x)$ is defined as a point $x_0$ such that $f(x) = 0$
  • A solution $x(t)$ of $x’ = f(x)$ is called an equilibrium solution if $x(t) = x_0$
  • For example, the critical points of $x’ = x(3 - x)$ are 0 and 3, so the equilibrium solutions are $x(t) = 3, x(t) = 5$

Slope Fields

  • Slope fields are drawn by calculating the slope at different points and drawing them
    • Given an equation $y’ = f(x, y)$, find the slope of y at various points and draw the slope as a line segment
  • Slope fields can be used to determine the shapes of solutions depending on an initial value
    • There are various functions that you can find from a slope field; how do we know if a function exists or is unique?

Picard’s Theorem

Definition. Consider the initial value problem y′=f(x,y),y(x0)=y0If the following two conditions are met:f(x,y) is continuous near (x0,y0)∂f∂y exists and is continuous near (x0,y0)then a solution to this IVP exists and is unique for (at least) some x near (x0,y0)\text{Definition. Consider the initial value problem } y' = f(x, y), y(x_0) = y_0 \\ \text{If the following two conditions are met:} \\ f(x, y) \text{ is continuous near } (x_0, y_0) \\ \frac{\partial f}{\partial y} \text{ exists and is continuous near } (x_0, y_0) \\ \text{then a solution to this IVP exists and is unique for (at least) some x near } (x_0, y_0)

Higher Order Linear ODEs

Inhomogenous

These functions take the form of ${a_ny^{(n)} + a_{n-1}y^{(n-1)} + … + a_1y’ + a_0y = f(x)}$. Note that you need one particular solution and one general solution, and the general solution can be found by setting the right hand side to 0; this degenerates the ODE to a homogenous equation.

Undetermined Coefficients

In order to find a particular solution, we can plug in different forms of solutions and find the coefficients that make the form true. These strategies involve adding factors of x such that the form is not a solution to the homogenous ODE.

f(x) is a polynomial

Given a d degree polynomial, the particular solution will take the form of ${y_p = (a_dx^d + … + a_1x + a_0)x^s}$ where s is the multiplicity of the root $r=0$.

f(x) is exponential

The particular solution will take the form of $y_p = ax^se^{kx}$, where s is the multiplicity of the root k in the characteristic equation and a is to be found.

f(x) is trigonometric

The particular solution will take the form of $y_p = (ae^{\alpha x}\cos \beta x + be^{\alpha x}\sin \beta x)x^s$, where s is the multiplicity of the complex root $\alpha \pm \beta i$.

Example:

y′′+y′=2cos⁡xyc=C1+C2e−xyp=acos⁡x+bsin⁡xyp′=−asin⁡x+bcos⁡xyp′′=−acos⁡x−bsin⁡xyp′′+yp′=(−a+b)cos⁡x+(−a−b)sin⁡x=2cos⁡x+0sin⁡xa=−1, b=1yp=−cos⁡x+sin⁡xy'' + y' = 2 \cos x \\ y_c = C_1 + C_2e^{-x} \\ y_p = a \cos x + b \sin x \\ y_p' = -a \sin x + b \cos x \\ y_p'' = -a \cos x - b \sin x \\ y_p'' + y_p' = (-a + b) \cos x + (-a - b)\sin x = 2 \cos x + 0 \sin x \\ a = -1\text{, }b = 1\\ y_p = - \cos x + \sin x

Combinations

If the right hand side is given by $f(x) + g(x)$ and $y_f$ solves $Ly = f(x)$ and $y_g$ solves $Ly = g(x)$, then $y_p = y_g + y_f$ is a particular solution.

Systems of ODEs

  • A system of differential equations is a finite set of diff. eqs.
  • Systems can obey different properties depending on whether or not all ODEs have that property; e.g. if all ODEs are linear, then the system is linear
    • The highest order in any ODE defines the order of the system
  • In this course, we will have the number of equations equal to the number of dependent variables

First-Order

  • Form: $y_1’ = g_1(y_1, y_2, \cdots, y_n, x), \cdots, y_n’ = g_n(y_1, y_2, \cdots, y_n, x)$
  • We can convert an ODE with one dependent of nth order to a system of ODEs with n dependent variables of first order
    • Ex. $y’’’ - y’’ + 2y = 1$ can be converted to ${u_3’ - u_3 + 2u_1 = 1\, u_2’ = u_3, u_1’ = u_2 }$, where ${u_3 = y’’}, {u_2 = y’}, {u_1 = y}$
    • General strategy: given $y^{(n)} = F(y^{(n-1)}, \cdots, y’, y, x)$, convert using $u_1 = y, u_2 = y’, \cdots, u_n = y^{(n-1)}$ and $u_1’ = u_2, \cdots, u_{n-1}’ = u_n, u_{n}’ = F(u_n, \cdots, u_1, x)$
    • Used to convert systems to nth order linear ODEs